(a^2+b^2+c^2) x 2 = 2 x (a^4+b^4+c^4)

suy ra: (a+b+c)^2 x 2 = (a+b+c)^4 x 2

Mà a+b+c= 0(gt)

suy ra: 0^2 x 2=0^4 x 2

0 = 0

=)))

\(\frac{a^4}{\left(a^2-b^2+c^2\right)\left(a^2+b^2-c^2\right)}=\frac{a^4}{\left[\left(a-b\right)\left(a+b\right)+c^2\right]\left[\left(a-c\right)\left(a+c\right)+b^2\right]}\)

\(\frac{a^4}{\left[-c\left(a-b\right)+c^2\right]\left[-b\left(a-c\right)+b^2\right]}=\frac{a^4}{4bc\left(b+c\right)^2}=\frac{a^4}{4a^2bc}\)

Tương tự với 2 phân thức còn lại, ta cũng có : \(\frac{b^4}{b^4-\left(c^2-a^2\right)^2}=\frac{b^4}{4ab^2c};\frac{c^4}{c^4-\left(a^2-b^2\right)^2}=\frac{c^4}{4abc^2}\)

\(VT=\frac{a^4}{4a^2bc}+\frac{b^4}{4ab^2c}+\frac{c^4}{4abc^2}=\frac{a^4bc+ab^4c+abc^4}{4a^2b^2c^2}=\frac{abc\left(a^3+b^3+c^3\right)}{4a^2b^2c^2}\)

\(VT=\frac{a^3+b^3+c^3}{4abc}\)

Mà \(a+b+c=0\) nên \(a^3+b^3+c^3=3abc\) ( tự cm ) 

\(\Rightarrow\)\(VT=\frac{3abc}{4abc}=\frac{3}{4}\) ( đpcm ) 

Chúc bạn học tốt ~ 

Đặt :

\(A=\)\(\dfrac{a^4}{a^4-\left(b^2-c^2\right)^2}+\dfrac{b^4}{b^4-\left(c^2-a^2\right)^2}+\dfrac{c^4}{c^4-\left(a^2-b^2\right)}\)

\(=\dfrac{a^4}{\left(a^2-b^2+c^2\right)\left(a^2+b^2-c^2\right)}+\dfrac{b^4}{\left(b^2-c^2+a^2\right)\left(b^2+c^2-a^2\right)}+\dfrac{c^4}{\left(c^2-a^2+b^2\right)\left(c^2+a^2-b^2\right)}\)

Ta có : \(a+b+c=0\)

\(\Leftrightarrow a+b=-c\)

\(\Leftrightarrow\left(a+b\right)^2=\left(-c\right)^2\)

\(\Leftrightarrow a^2+2ab+b^2=c^2\)

\(\Leftrightarrow a^2+b^2-c^2=-2ab\)

Tương tự :

+) \(a^2-b^2+c^2=-2ac\)

+) \(b^2+c^2-a^2=-2bc\)

\(\Leftrightarrow A=\dfrac{a^4}{\left(-2ac\right)\left(-2ab\right)}+\dfrac{b^4}{\left(-2ab\right)\left(-2bc\right)}+\dfrac{c^4}{\left(-2bc\right)\left(-2ac\right)}\)

\(=\dfrac{a^4}{4a^2bc}+\dfrac{b^4}{4ab^2c}+\dfrac{c^4}{4abc^2}\)

\(=\dfrac{a^4bc+ab^4c+abc^4}{4a^2b^2c^2}\)

\(=\dfrac{abc\left(a^3+b^3+c^3\right)}{4a^2b^2c^2}\) (cậu tự chứng minh \(a^3+b^3+c^3=3abc\) nhé)

\(=\dfrac{3a^2b^2c^2}{4a^2b^2c^2}\)

\(=\dfrac{3}{4}\)

Vậy..